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A085697
a(n) = T(n)^2, where T(n) = A000073(n) is the n-th tribonacci number.
17
0, 0, 1, 1, 4, 16, 49, 169, 576, 1936, 6561, 22201, 75076, 254016, 859329, 2907025, 9834496, 33269824, 112550881, 380757169, 1288092100, 4357584144, 14741602225, 49870482489, 168710633536, 570743986576, 1930813074369, 6531893843049
OFFSET
0,5
COMMENTS
In general, squaring the terms of a third-order linear recurrence with signature (x,y,z) will result in a sixth-order recurrence with signature (x^2 + y, x^2*y + z*x + y^2, x^3*z + 4*x*y*z - y^3 + 2*z^2, x^2*z^2 - x*y^2*z - z^2*y, z^2*y^2 - z^3*x, -z^4). - Gary Detlefs, Jan 10 2023
REFERENCES
R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.
FORMULA
G.f.: x^2*( 1-x-x^2-x^3 )/( (1-3*x-x^2-x^3)*(1+x+x^2-x^3) ).
a(n+6) = 2*a(n+5) + 3*a(n+4) + 6*a(n+3) - a(n+2) - a(n).
a(n) = (-A057597(n-2) + 3*A057597(n-1) + 6*A057597(n) + 5*A113300(n-1) - A099463(n-2))/11. - R. J. Mathar, Aug 19 2008
MATHEMATICA
LinearRecurrence[{2, 3, 6, -1, 0, -1}, {0, 0, 1, 1, 4, 16}, 30] (* Harvey P. Dale, Oct 26 2020 *)
PROG
(Maxima)
t[0]:0$ t[1]:0$ t[2]:1$
t[n]:=t[n-1]+t[n-2]+t[n-3]$
makelist(t[n]^2, n, 0, 40); /* Emanuele Munarini, Mar 01 2011 */
(Magma)
R<x>:=PowerSeriesRing(Integers(), 40); [0, 0] cat Coefficients(R!( x^2*(1-x-x^2-x^3)/((1-3*x-x^2-x^3)*(1+x+x^2-x^3)) )); // G. C. Greubel, Nov 20 2021
(Sage)
@CachedFunction
def T(n): # A000073
if (n<2): return 0
elif (n==2): return 1
else: return T(n-1) +T(n-2) +T(n-3)
def A085697(n): return T(n)^2
[A085697(n) for n in (0..40)] # G. C. Greubel, Nov 20 2021
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Emanuele Munarini, Jul 18 2003
EXTENSIONS
Offset corrected to match A000073 by N. J. A. Sloane, Sep 12 2020
Name corrected to match corrected offset by Michael A. Allen, Jun 10 2021
STATUS
approved